Circuit Analysis of 1st Order RC Circuit

Hello

Please the attached image below. I want to know at t<0 when the switch is closed for a long time as question describes then how to find the voltage i.e. v(0-) across the capacitor?
where,
v(0-) is the voltage across the capacitor just before opening the switch at t=0



PS: We know capacitor voltage doesn't change instantaneously so v(0-)=v(0+)=v(0)

Best Regards,
Princess

Well since the problem says assume the switch has been open for a long time you are supposed to assume steady state conditions meaning the capacitor is fully charged. In steady state when t->infinity a capacitor looks like an open circuit to a DC voltage. Therefore there was no current flowing anymore, no voltage drop across the resistor and therefore 10V across the cap.

No, you are not understanding. when t<0 the switch is open. At t=0 the switch is closed.
When t<0, then vc=10V.

But when t>0 then switch is closed and now two voltage sources will contribute to the capacitor voltage. This is the problem for what I want to know. How to solve for capacitor voltage for t>0?

I hope you have understood.

Ah, well the original post said

how to find the voltage i.e. v(0-) across the capacitor?

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Anyway once the switch closes the capacitor will begin to decay from the value of V(0-). A good place to start is by finding what the steady state voltage will be when t approaches infinity and the cap is finished discharging. Since again we know a cap looks like an open circuit at steady state to a DC source this is a simple voltage divider between the two resistors and the two voltage sources.

Vc(t=inf) = (6/(6+2))*15-5 = 6.25V

Then you need to factor in the exponential decay rate of the capacitor based on the RC time constant tau. The time constant will be based on the capacitance of 1/3 F and the parallel combination of the two resistors 1.5 Ohm. This gives a time constant of 0.5 seconds. Therefore the capacitor will discharge down to it steady state value of 6.25V exponentially with e^(-t/RC). As for the 3.75V it simply comes from the total amount that the voltage will decay. Hope this helps.

Where does this 15-5 come?

So normally you see a resistor divider as R2/(R1+R2) times the source voltage to get the voltage in between the two resistors. Often one side of the divider is ground so it works out fine, but in this case you have two different potential levels on either side of the divider. Therefore you have to multiply the resistor ratio by the entire voltage drop from end to end which in this case is 15V. Maybe theres a better way to think about it but I basically pretend that the source on the left is +15V and the source on the right is grounded. Then you can solve the voltage divider the normal way and subtract 5 at the end to account for the change you made.

download ltspice and look at the waveforms..then you understand theory much better...your university wont tell you that. For some reason they hate simulators. and want students to work it out longhand and suffer.

treez said:

.....your university wont tell you that. For some reason they hate simulators. and want students to work it out longhand and suffer.

Click to expand...

Your experients? In general - not true.

in English universities it is true.....I know in Germany you rightly facilitate good learning for your students...Hence you have hundreds more engineering companies. Hence you have Porsche, BMW, Volkeswagen, Audi, Mercedes and in England, we have the following cars that we design...errrr...errr.....I cant think of any....come to think of it there aren't any......now lets think of some big English companies that do electronics.....errrr.....errrr......Dyson......that's it

@KD494
Sorry didn't understand yet how to get 15-5